Flamant solution

Flamant solution

The Flamant solution provides expressions for the stresses and displacements in a linear elastic wedge loaded by point forces at its sharp end. This solution was developed by A. Flamant [A. Flamant. (1892). "Sur la répartition des pressions dans un solide rectangulaire chargé transversalement." Compte. Rendu. Acad. Sci. Paris, vol. 114, p. 1465.] in 1892 by modifying the three-dimensional solution of Boussinesq.

The stresses predicted by the Flamant solution are (in polar coordinates): egin{align} sigma_{rr} & = frac{2C_1cos heta}{r} + frac{2C_3sin heta}{r} \ sigma_{r heta} & = 0 \ sigma_{ heta heta} & = 0 end{align}where C_1, C_3 are constants that are determined from the boundary conditions and the geometry of the wedge (i.e., the angles alpha,eta) and satisfy: egin{align} F_1 & + 2int_{alpha}^{eta} (C_1cos heta + C_3sin heta)~cos heta~ d heta = 0 \ F_2 & + 2int_{alpha}^{eta} (C_1cos heta + C_3sin heta)~sin heta~ d heta = 0 end{align}where F_1,F_2 are the applied forces. The wedge problem is "self-similar" and has no inherent length scale. Also, all quantities can be expressed in the separated-variable form sigma = f(r)g( heta). The stresses vary as (1/r).

Forces acting on a half-plane

For the special case where alpha = -pi, eta = 0, the wedge is converted into a half-plane with a normal force and a tangential force. In that case:C_1 = - frac{F_1}{pi} ~;~~ C_3 = -frac{F_2}{pi}Therefore the stresses are: egin{align} sigma_{rr} & = -frac{2}{pi~r} (F_1~cos heta + F_2~sin heta) \ sigma_{r heta} & = 0 \ sigma_{ heta heta} & = 0 end{align}and the displacements are (using Michell's solution): egin{align} u_r & = -cfrac{1}{4pimu}left [F_1{(kappa-1) hetasin heta - cos heta + (kappa+1)ln rcos heta} + ight. \ & qquad qquad left. F_2{(kappa-1) hetacos heta + sin heta - (kappa+1)ln rsin heta} ight] \ u_ heta & = -cfrac{1}{4pimu}left [F_1{(kappa-1) hetacos heta - sin heta - (kappa+1)ln rsin heta} - ight. \ & qquad qquad left. F_2{(kappa-1) hetasin heta + cos heta + (kappa+1)ln rcos heta} ight] end{align} The ln r dependence of the displacements implies that the displacement grows the further one moves from the point of application of the force (and is unbounded at infinity). This feature of the Flamant solution is confusing and appears unphysical. For a discussion of the issue see [http://imechanica.org/node/319 http://imechanica.org/node/319] .

Displacements at the surface of the half-plane

The displacements in the x_1, x_2 directions at the surface of the half-plane are given by:egin{align}u_1 & = frac{F_1(kappa+1)ln|x_1{4pimu} +frac{F_2(kappa+1) ext{sign}(x_1)}{8mu} \u_2 & = frac{F_2(kappa+1)ln|x_1{4pimu} +frac{F_1(kappa+1) ext{sign}(x_1)}{8mu} end{align}where:kappa = egin{cases} 3 - 4 u & qquad ext{plane strain} \ cfrac{3 - u}{1+ u} & qquad ext{plane stress} end{cases} u is the Poisson's ratio, mu is the shear modulus, and: ext{sign}(x) = egin{cases} +1 & x > 0 \ -1 & x < 0 end{cases}

Derivation of Flamant solution

If we assume the stresses to vary as (1/r), we can pick terms containing 1/r in the stresses from Michell's solution. Then the Airy stress function can be expressed as:varphi = C_1 r hetasin heta + C_2 rln r cos heta + C_3 r hetacos heta + C_4 rln r sin heta Therefore, from the tables in Michell's solution, we have:egin{align}sigma_{rr} & = C_1left(frac{2cos heta}{r} ight) + C_2left(frac{cos heta}{r} ight) + C_3left(frac{2sin heta}{r} ight) + C_4left(frac{sin heta}{r} ight) \sigma_{r heta} & = C_2left(frac{sin heta}{r} ight) + C_4left(frac{-cos heta}{r} ight) \sigma_{ heta heta} & = C_2left(frac{cos heta}{r} ight) + C_4left(frac{sin heta}{r} ight) end{align}The constants C_1, C_2, C_3, C_4 can then, in principle, be determined from the wedge geometry and the applied boundary conditions.

However, the concentrated loads at the vertex are difficult to express in terms of traction boundary conditions because
# the unit outward normal at the vertex is undefined
# the forces are applied at a point (which has zero area) and hence the traction at that point is infinite.

To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge. [Slaughter, W. S. (2002). "The Linearized Theory of Elasticity". Birkhauser, Boston, p. 294.] [ J. R. Barber, 2002, "Elasticity: 2nd Edition", Kluwer Academic Publishers. ] Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius a,. Along the arc of the circle, the unit outward normal is mathbf{n} = mathbf{e}_r where the basis vectors are (mathbf{e}_r, mathbf{e}_ heta). The tractions on the arc are: mathbf{t} = oldsymbol{sigma}cdotmathbf{n} quad implies t_r = sigma_{rr}, ~ t_ heta = sigma_{r heta} ~.

Next, we examine the force and moment equilibrium in the bounded wedge and get: egin{align} sum f_1 & = F_1 + int_{alpha}^{eta} left [sigma_{rr}(a, heta)~cos heta - sigma_{r heta}(a, heta)~sin heta ight] ~a~d heta = 0 \ sum f_2 & = F_2 + int_{alpha}^{eta} left [sigma_{rr}(a, heta)~sin heta + sigma_{r heta}(a, heta)~cos heta ight] ~a~d heta = 0 \ sum m_3 & = int_{alpha}^{eta} left [a~sigma_{r heta}(a, heta) ight] ~a~d heta = 0 end{align} We require that these equations be satisfied for all values of a, and thereby satisfy the boundary conditions.

The traction-free boundary conditions on the edges heta=alpha and heta=eta also imply that: sigma_{r heta} = sigma_{ heta heta} = 0 qquad ext{at}~~ heta=alpha, heta=eta except at the point r = 0.

If we assume that sigma_{r heta}=0 everywhere, then the traction-free conditions and the moment equilibrium equation are satisfied and we are left with : egin{align} F_1 & + int_{alpha}^{eta} sigma_{rr}(a, heta)~a~cos heta ~d heta = 0 \ F_2 & + int_{alpha}^{eta} sigma_{rr}(a, heta)~a~sin heta ~d heta = 0 end{align} and sigma_{ heta heta} = 0 along heta=alpha, heta=eta except at the point r = 0. But the field sigma_{ heta heta} = 0 everywhere also satisfies the force equilibrium equations. Hence this must be the solution. Also, the assumption sigma_{r heta}=0 implies that C_2 = C_4 = 0.

Therefore,:sigma_{rr} = frac{2C_1cos heta}{r} + frac{2C_3sin heta}{r} ~;~~sigma_{r heta} = 0 ~;~~ sigma_{ heta heta} = 0

To find a particular solution for sigma_{rr} we have to plug in the expression for sigma_{rr} into the force equilibrium equations to get a system of two equations which have to be solved for C_1, C_3:: egin{align} F_1 & + 2int_{alpha}^{eta} (C_1cos heta + C_3sin heta)~cos heta~ d heta = 0 \ F_2 & + 2int_{alpha}^{eta} (C_1cos heta + C_3sin heta)~sin heta~ d heta = 0 end{align}

Forces acting on a half-plane

If we take alpha = -pi and eta = 0, the problem is converted into one where a normal force F_2 and a tangential force F_1 act on a half-plane. In that case, the force equilibrium equations take the form: egin{align} F_1 & + 2int_{-pi}^{0} (C_1cos heta + C_3sin heta)~cos heta~ d heta = 0 qquad implies F_1 + C_1pi = 0\ F_2 & + 2int_{-pi}^{0} (C_1cos heta + C_3sin heta)~sin heta~ d heta = 0 qquad implies F_2 + C_3pi = 0 end{align} Therefore: C_1 = - cfrac{F_1}{pi} ~;~~ C_3 = - cfrac{F_2}{pi} ~. The stresses for this situation are: sigma_{rr} = -frac{2}{pi r}(F_1cos heta + F_2sin heta) ~;~~sigma_{r heta} = 0 ~;~~ sigma_{ heta heta} = 0 Using the displacement tables from the Michell solution, the displacements for this case are given by: egin{align} u_r & = -cfrac{1}{4pimu}left [F_1{(kappa-1) hetasin heta - cos heta + (kappa+1)ln rcos heta} + ight. \ & qquad qquad left. F_2{(kappa-1) hetacos heta + sin heta - (kappa+1)ln rsin heta} ight] \ u_ heta & = -cfrac{1}{4pimu}left [F_1{(kappa-1) hetacos heta - sin heta - (kappa+1)ln rsin heta} - ight. \ & qquad qquad left. F_2{(kappa-1) hetasin heta + cos heta + (kappa+1)ln rcos heta} ight] end{align}

Displacements at the surface of the half-plane

To find expressions for the displacements at the surface of the half plane, we first find the displacements for positive x_1 ( heta=0) and negative x_1 ( heta = pi) keeping in mind that r = |x_1| along these locations.

For heta=0 we have: egin{align} u_r = u_1 & = cfrac{F_1}{4pimu}left [1 - (kappa+1)ln |x_1| ight] \ u_ heta = u_2 & = cfrac{F_2}{4pimu}left [1 + (kappa+1)ln |x_1| ight] end{align} For heta = pi we have: egin{align} u_r = -u_1 & = -cfrac{F_1}{4pimu}left [1 - (kappa+1)ln |x_1| ight] + cfrac{F_2}{4mu}(kappa-1)\ u_ heta = -u_2 & = cfrac{F_1}{4mu}(kappa-1) - cfrac{F_2}{4pimu}left [1 + (kappa+1)ln |x_1| ight] end{align} We can make the displacements symmetric around the point of application of the force by adding rigid body displacements (which does not affect the stresses) : u_1 = cfrac{F_2}{8mu}(kappa-1) ~;~~ u_2 = cfrac{F_1}{8mu}(kappa-1) and removing the redundant rigid body displacements: u_1 = cfrac{F_1}{4pimu} ~;~~ u_2 = cfrac{F_2}{4pimu} ~. Then the displacements at the surface can be combined and take the form: egin{align} u_1 & = cfrac{F_1}{4pimu}(kappa+1)ln |x_1| + cfrac{F_2}{8mu}(kappa-1) ext{sign}(x_1) \ u_2 & = cfrac{F_2}{4pimu}(kappa+1)ln |x_1| +cfrac{F_1}{8mu}(kappa-1) ext{sign}(x_1) end{align} where: ext{sign}(x) = egin{cases} +1 & x > 0 \ -1 & x < 0 end{cases}

References

See also

* Michell solution
* Linear elasticity
* Stress (physics)


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